# Prove that $a^ab^b>\left(\frac{a+b}{2}\right)^{a+b}$ where $a\ne b$

Prove that $$a^a\cdot b^b>\left(\dfrac{a+b}{2}\right)^{a+b}$$ where $$a\ne b$$.

My work:

$$a^a\cdot b^b>\left(\frac{a+b}{2}\right)^a\cdot\left(\frac{a+b}{2}\right)^b\implies 1>\left(\frac{1+\frac{b}{a}}{2}\right)^a\cdot\left(\frac{1+\frac{a}{b}}{2}\right)^b$$

Take the function $f(x)=x\ln x,$ where $x>0$, then $f'(x)=1+\ln(x)$ and $f''(x)=\frac{1}{x}>0,$ so $f(x)$ is convex functon. Apply jensen-inequality in $f$, we have $$\dfrac{a\ln a + b\ln b}{2}\ge \dfrac{(a+b)}{2}\ln\bigg(\dfrac{a+b}{2}\bigg),$$
on simplification and taking anti-log on both sides it becomes, $a^ab^b\ge \bigg(\dfrac{a+b}{2}\bigg)^{a+b}$.
• the exponent on the rhs of the original equation is $a+b$, however - is that a typo or does the claim still hold ? – collapsar Feb 25 '14 at 11:14
You have to reason with logarithms. Firstly you have to know that (property of convexity) $$\ln\left(\frac{\sum x_i}{n}\right)<\frac{\sum\ln(x_i)}{n}.$$
it also holds that the logarithm of a simple average is less than the weighted average of logarithms, i.e.: $$\ln\left(\frac{\sum x_i}{n}\right)<\sum w_i\ln(x_i), \sum w_i=1.$$ In your case $$w_1=\frac{a}{a+b}, w_2=\frac{b}{a+b},$$ take the exponential and you obtain the thesis.